$g(n) = 7n^{3}-2n^{2}+6n-h(n)$ $h(x) = 2x^{2}-7x$ $f(t) = 4t^{2}-4t-3(h(t))$ $ f(h(4)) = {?} $
First, let's solve for the value of the inner function, $h(4)$ . Then we'll know what to plug into the outer function. $h(4) = 2(4^{2})+(-7)(4)$ $h(4) = 4$ Now we know that $h(4) = 4$ . Let's solve for $f(h(4))$ , which is $f(4)$ $f(4) = 4(4^{2})+(-4)(4)-3(h(4))$ To solve for the value of $f$ , we need to solve for the value of $h(4)$ $h(4) = 2(4^{2})+(-7)(4)$ $h(4) = 4$ That means $f(4) = 4(4^{2})+(-4)(4)+(-3)(4)$ $f(4) = 36$